Q.Find whether the function is continuous or discontinuous at the indicated point: $f(x) = \begin{cases} \dfrac{e^{1/x}}{1 + e^{1/x}}, & x \ne 0 \\ 0, & x = 0 \end{cases}$ at $x = 0$.

Concept: Continuity At A Point — A function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. For piecewise functions, check the left-hand limit, right-hand limit, and the function value.

Step 1: Evaluate $f(0)$

Given directly: $f(0) = 0$.

Step 2: Compute $\lim_{x \to 0^+} f(x)$

As $x \to 0^+$, $\frac{1}{x} \to +\infty$, so $e^{1/x} \to \infty$.

Then $f(x) = \frac{e^{1/x}}{1 + e^{1/x}} \to \frac{\infty}{\infty} \to 1$ (divide numerator and denominator by $e^{1/x}$: $\frac{1}{e^{-1/x} + 1} \to 1$).

Thus $\lim_{x \to 0^+} f(x) = 1$.

Step 3: Compute $\lim_{x \to 0^-} f(x)$

As $x \to 0^-$, $\frac{1}{x} \to -\infty$, so $e^{1/x} \to 0$. …

The Core Idea: Continuity at a Point

A function is continuous at a point $x = a$ if three things hold:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

Here, $f(0) = 0$ is given, so condition 1 is satisfied. The real question is whether the limit exists. The function involves $e^{1/x}$, which behaves very differently depending on whether $x$ approaches 0 from the left ($x \to 0^-$) or from the right ($x \to 0^+$). This is because $1/x$ shoots to $-\infty$ on the left and $+\infty$ on the right. So we must check the one-sided limits separately.

Step-by-Step Solution

1. Understand the behaviour of $e^{1/x}$ near $x=0$.

When $x$ is a small positive number (say $x = 0.001$), $1/x$ is a large positive number, so $e^{1/x}$ becomes enormous. As $x \to 0^+$, $1/x \to +\infty$, hence $e^{1/x} \to +\infty$.

When $x$ is a small negative number (say $x = -0.001$), $1/x$ is a large negative number, so $e^{1/x}$ becomes very close to 0. As $x \to 0^-$, $1/x \to -\infty$, hence $e^{1/x} \to 0$.

This difference is the key to the entire problem.

2. Compute the right-hand limit ($x \to 0^+$).

We need $\displaystyle \lim_{x \to 0^+} \frac{e^{1/x}}{1 + e^{1/x}}$.

Since $e^{1/x} \to \infty$, both numerator and denominator blow up. A standard trick is to divide the numerator and denominator by $e^{1/x}$:

Now as $x \to 0^+$, $e^{-1/x} \to e^{-\infty} = 0$. So the expression approaches $\frac{1}{0 + 1} = 1$.

Thus, $\displaystyle \lim_{x \to 0^+} f(x) = 1$.

3. Compute the left-hand limit ($x \to 0^-$).

We need $\displaystyle \lim_{x \to 0^-} \frac{e^{1/x}}{1 + e^{1/x}}$.

Here, $e^{1/x} \to 0$. So the numerator tends to 0 and the denominator tends to $1 + 0 = 1$. Therefore, the whole fraction tends to $0/1 = 0$.

Thus, $\displaystyle \lim_{x \to 0^-} f(x) = 0$. …

Method: Testing Continuity via One-Sided Limits of an Exponential Expression

This method applies whenever a function involves $e^{1/x}$ (or any $e^{g(x)}$ where $g(x) \to +\infty$ from one side and $g(x)\to -\infty$ from the other side of the point being tested) — the exponential's behaviour is fundamentally different on the two sides, so a single substitution is never valid here.

Steps

Step 1: Analyse how the inner expression behaves on each side of the point.

For $\frac1x$ at $x=0$: as $x \to 0^+$, $\frac1x \to +\infty$; as $x \to 0^-$, $\frac1x \to -\infty$. This asymmetry is the whole reason one-sided limits are unavoidable.

Step 2: Translate that into exponential behaviour.

So $e^{1/x} \to \infty$ as $x\to 0^+$ and $e^{1/x} \to 0$ as $x \to 0^-$.

Step 3: Handle the $\infty/\infty$ side by dividing numerator and denominator by the dominant exponential term. …

Common Mistakes

Mistake 1: Evaluating the limit as if $e^{1/x}$ behaves the same way from both sides of $x=0$.

Why it's wrong: $\frac1x \to +\infty$ as $x\to 0^+$ but $\frac1x \to -\infty$ as $x \to 0^-$, so $e^{1/x}$ blows up on one side and vanishes on the other — treating them identically (e.g. substituting a single limit value for $e^{1/x}$ everywhere) produces a wrong, and often self-contradictory, result. Correct approach: always split into the right-hand limit ($x\to0^+$) and left-hand limit ($x\to0^-$) and evaluate each with its own behaviour of $e^{1/x}$.

Mistake 2: Leaving the $\infty/\infty$ side unresolved instead of dividing through by the dominant exponential term. …